3.2.13 \(\int \frac {3+x^2}{(1+x^2)^2} \, dx\) [113]

Optimal. Leaf size=14 \[ \frac {x}{1+x^2}+2 \tan ^{-1}(x) \]

[Out]

x/(x^2+1)+2*arctan(x)

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Rubi [A]
time = 0.00, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {393, 209} \begin {gather*} 2 \text {ArcTan}(x)+\frac {x}{x^2+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3 + x^2)/(1 + x^2)^2,x]

[Out]

x/(1 + x^2) + 2*ArcTan[x]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 393

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*c - a*d))*x*((a + b*x^n)^(p
+ 1)/(a*b*n*(p + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x]
/; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rubi steps

\begin {align*} \int \frac {3+x^2}{\left (1+x^2\right )^2} \, dx &=\frac {x}{1+x^2}+2 \int \frac {1}{1+x^2} \, dx\\ &=\frac {x}{1+x^2}+2 \tan ^{-1}(x)\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 14, normalized size = 1.00 \begin {gather*} \frac {x}{1+x^2}+2 \tan ^{-1}(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 + x^2)/(1 + x^2)^2,x]

[Out]

x/(1 + x^2) + 2*ArcTan[x]

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Maple [A]
time = 0.07, size = 15, normalized size = 1.07

method	result	size
default	\(\frac {x}{x^{2}+1}+2 \arctan \left (x \right )\)	\(15\)
risch	\(\frac {x}{x^{2}+1}+2 \arctan \left (x \right )\)	\(15\)
meijerg	\(-\frac {x}{2 \left (x^{2}+1\right )}+2 \arctan \left (x \right )+\frac {3 x}{2 x^{2}+2}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2+3)/(x^2+1)^2,x,method=_RETURNVERBOSE)

[Out]

x/(x^2+1)+2*arctan(x)

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Maxima [A]
time = 0.49, size = 14, normalized size = 1.00 \begin {gather*} \frac {x}{x^{2} + 1} + 2 \, \arctan \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+3)/(x^2+1)^2,x, algorithm="maxima")

[Out]

x/(x^2 + 1) + 2*arctan(x)

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Fricas [A]
time = 1.18, size = 19, normalized size = 1.36 \begin {gather*} \frac {2 \, {\left (x^{2} + 1\right )} \arctan \left (x\right ) + x}{x^{2} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+3)/(x^2+1)^2,x, algorithm="fricas")

[Out]

(2*(x^2 + 1)*arctan(x) + x)/(x^2 + 1)

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Sympy [A]
time = 0.03, size = 10, normalized size = 0.71 \begin {gather*} \frac {x}{x^{2} + 1} + 2 \operatorname {atan}{\left (x \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2+3)/(x**2+1)**2,x)

[Out]

x/(x**2 + 1) + 2*atan(x)

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Giac [A]
time = 1.20, size = 14, normalized size = 1.00 \begin {gather*} \frac {x}{x^{2} + 1} + 2 \, \arctan \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+3)/(x^2+1)^2,x, algorithm="giac")

[Out]

x/(x^2 + 1) + 2*arctan(x)

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Mupad [B]
time = 0.01, size = 14, normalized size = 1.00 \begin {gather*} 2\,\mathrm {atan}\left (x\right )+\frac {x}{x^2+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2 + 3)/(x^2 + 1)^2,x)

[Out]

2*atan(x) + x/(x^2 + 1)

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